Question

current attempt in progress
the drawing shows a block at rest on an incline. the mass of the block is 8.0 kg. what is the static frictional force that acts on the block?
73 n
32 n
78 n
29 n
there is not enough information to calculate the frictional force.

Explanation:

Step1: Analyze forces on block

The block is at rest on the incline. The gravitational - force component along the incline is balanced by the static frictional force. The gravitational force is $F_g = mg$, and the component of the gravitational force along the incline is $F_{g\parallel}=mg\sin\theta$, where $m = 8.0\ kg$, $g = 9.8\ m/s^2$, and $\theta=22^{\circ}$.

Step2: Calculate the gravitational - force component along the incline

$F_{g\parallel}=mg\sin\theta=8.0\times9.8\times\sin(22^{\circ})$.
We know that $\sin(22^{\circ})\approx0.375$, so $F_{g\parallel}=8.0\times9.8\times0.375 = 8\times9.8\times\frac{3}{8}=29.4\ N\approx29\ N$. Since the block is at rest, the static frictional force $f_s$ balances $F_{g\parallel}$, so $f_s = F_{g\parallel}$.

Answer:

29 N